channels are not queues, as Justin said
On Saturday, April 20, 2024 at 8:18:18 PM UTC-4 Justin Israel wrote:
> On Sunday, April 21, 2024 at 11:18:24 AM UTC+12 Taňryberdi Şyhmyradow
> wrote:
>
> Hello guys,
> For the following lines, I wanted to print numbers in ordered, but
> couldn't. Could you please help me and explain the reason
> Thanks in advance
>
> ```
> numbers := []int{1, 2, 3, 4, 5}
>
> // Create a buffered channel to handle multiple values
> printed := make(chan int, len(numbers))
>
> for _, n := range numbers {
> fmt.Println("Sending", n, "to the channel")
> go func() {
> printed <- n
> }() // Pass the value of n by copying it
> }
>
> // Receive all values from the channel in a loop
> for i := 0; i < len(numbers); i++ {
> fmt.Println(<-printed)
> }
> ```
>
>
> When you start a bunch of goroutines in a loop, there is no guarantee as
> to what order the scheduler will start each one.
> Thus you will see them delivered in different orders on each run. You have
> to decide on some form of synchronization. Maybe you choose to run a single
> goroutine worker that will loop over the source slice, and push the values
> into the channel in order. Or maybe you will keep using many goroutines but
> collect them all in the receiver, sort them after the last value is
> received, and then print them out. Or, maybe your receiver will have some
> kind of buffering where it collects values and only prints them when it has
> the next one in sequence.
>
>
>
> --
> Tanryberdi Shyhmyradov
>
>
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